Where $A,B,C, k ,m,v_0,x_0,g$ are all constants and $t$ is the independent variable. I cannot figure out how to reconcile the two general solution and rewrite one as the other. There is no trigonmetric identity to relate the two and I am sure they must be the same expression as they plot very similarly. Can anyone provide some indications as to how I can rewrite the first form as the second form of the solution or vice versa?

$\begingroup$You cannot reconcile your two solutions $$ x(t) = A {\cos ({\sqrt {\frac k m}}t}) +B{\sin ({\sqrt {\frac k m}}t}) +{\frac {mg} k} $$ $$x(t) = C{\sin ({\sqrt{\frac k m}}}t+{\arctan({\sqrt{\frac k m}{\frac {x_0} {v_0}}}}))+{\frac {mg} k}$$ because the first is general (they are made of two independent fonctions of $t$) while the second isn't the general solution (one function of $t$ only). Moreover, the second is false : if you put it into the ODE, you observe that it doesn't agree. Probably a typo or a mistake. Why don't you show what you have done ?$\endgroup$
– JJacquelinDec 11 '16 at 18:22

$\begingroup$I've updated the original question to show the work I've done. Thanks.$\endgroup$
– user32882Dec 12 '16 at 7:40

1

$\begingroup$@JJacquelin Consider the equation of harmonic oscillator $\ddot{x} + \omega^2 x = 0$. Its general solution is $x(t) = C_1 \cos(\omega t) + C_2 \sin (\omega t) $ which also can be rewritten as $x(t) = \mathcal{A} \sin (\omega t + \varphi)$ :) The second independent function just hides in this form which is exactly what OP has.$\endgroup$
– EvgenyDec 12 '16 at 15:38